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最新推荐文章于 2025-03-30 00:23:39 发布

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本文介绍了一种快速计算奇数序列平方和的方法，并提供了一个简洁的C语言程序实现。通过数学推导，得出了一般化的奇数平方和公式。

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Calculate the formula

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5099 Accepted Submission(s): 1536

Problem Description

You just need to calculate the sum of the formula: 1^2+3^2+5^2+……+ n ^2.

Input

In each case, there is an odd positive integer n.

Output

Print the sum. Make sure the sum will not exceed 2^31-1

Sample Input

Sample Output

Author

wangye

Source

HDU 2007-11 Programming Contest_WarmUp

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威士忌

//已知:1^2+2^2+3^2+……+n^2 =n(n+1)(2n+1)/6 —①
//那么1^2+2^2+3^2+……+n^2+……+(2n+1)^2 =(2n+1)(n+1)(4n+3)/3 —②
//又有2^2+4^2+6^2+……+(2n)^2 =4[1^2+2^2+3^2+……+n^2]=4*①=2n(n+1)(2n+1)/3 —③
//设所求为S 比较②和③可知 S=②-③=(2n+1)(n+1)(4n+3)/3-2n(n+1)(2n+1)/3
//=(2n+1)(n+1)(2n+3)/3 —④
//因为S是2n+1项的和 把它一般化 则奇数项平方和一般公式Sn=n(n+1)(n+2)/6
#include <stdio.h>
#include <math.h>
int main()
{
	__int64 n;
	while(~scanf("%I64d", &n))
	{
		printf("%I64d\n", n * (n+1) * (n+2) / 6);
	}
	return 0;
}